CPhill

Thanks, Melody!!!!

We still arrived at the same station.....that's encouraging  !!!!

                        x^4  + 2x^3  + 5x^2   + 12x  + 29

x^2 - 2x - 1   [   x^6  +  0x^5  + 0x^4 + 0x^3 + 0x^2 + ax + b ]

                        x^6   -  2x^5   - x^4               

                      ________________________

                                 2x^5    + x^4  + 0x^3

                                 2x^5   -  4x^4 -  2x^3

                               _______________________

                                            5x^4 + 2x^3  + 0x^2

                                            5x^4 -10x^3  - 5x^2

                                           ___________________

                                                     12x^3 + 5x^2 + ax

                                                      12x^3 -24x^2 -12x

                                                    __________________________

                                                                 29x^2 + ( a + 12)x + b

                                                                 29x^2  -    58x      -29

                                                               ____________________________

                                                                              (a + 12 + 58)x + (b + 29)

(a + 70)x  + (b + 29)  =  0

ax + 70x + b + 29  =  0

ax + b   =   -70x - 29

So

a + b     =  -70 + (-29)     =  -99

"Amplitude"  does not apply to the tangent, cotangent, secant or cosecant functions

Thanks to EP for pointing this out to me   !!!!

1) Ticket cost for 60 passengers  =  4000 - (60 - 50)(50)  =   4000 - 10(50)  = R3500

2)  Price of ticket for over 50  =   R [  4000 - 50x  ] 

3) The revenue, R  = number of passengers  * cost of a ticket 

(50 + x)(4000 - 50x)  =   20000 + 4000x - 2500x - 50x^2   = 200000 + 1500x - 50x^2

4)   In the form    ax^2  + bx + c

The number of passengers that will provide the max revenue  is  -b/[2a]

So we have

R  = -50x^2 + 1500x + 200000

So......the number of passengers providing the max revenue is  : 

(-1500) / (2 * -50)  = -1500/ -100 = 15 passengers =  15 over 50  =  65 passengers

EDIT TO CORRECT A SILLY ERROR !!!

Let the first term =  2000

Let the fifth term = 10125

The exponetial form is

A = A₀ b^(t)        where  A₀  is the first term, t is the month [ Jan = 0 ]  and A is the amount in the "t" month

So.....we want to solve this  for b

10125 = 2000b^(4)

10125 = 2000b^4     divide both sides by 2000

10125 /2000 = b^4      take the 4th root of both sides

1.5  = b = growth factor

Jan  = 2000

Feb  = 2000(1.5) = 3500

Mar = 2000(1.5)^2 = 4500

Apr = 2000(1.5)^3 = 6750

May = 2000(1.5)^4  = 10125

We know that it  is exponential because we have a growth factor being raised to a power  and because the growth is not linear

Excellent, heureka!!!!

Excellent, FL!!!!!

P(sunshine 3 days in a row)  = (.75)^3  ≈ 0.42 ≈  42%

P (rain at least once during the week)  = 

1 - P( of no rain during the week ) =

1 - (.75)^ 7  ≈ .87 ≈  87%

(a)  If independent   

P(winning and playing at home) = P(winning) * P(playing at home)     ????

             0.2                                 =   0.25         *      0.8       ????

             0.2                                 =   0.2

So....they are independent

(b)   

P ( losing and playing away)   =  P (losing ) * P (playing away)   ????

           0.1                               =    0.7           *    0.15      ????

           0.1                              =     0.105

So....not independent